Assuming each solution to be 0.10 m , rank the following aqueous solutions in order of decreasing ph.

rank the solutions from the highest to lowest ph. to rank items as equivalent, overlap them

a. n2h2
b. ba(oh)2
c. hocld. naoh
e. hcl

Ba(OH)₂ > NaOH > HCl = HOCl >  N₂H₂

Explanation:

The pH = -log[H⁺] and pOH = -log[OH⁻], and by the water equilibrium, pH + pOH = 14. The value of pH determines the acid-base solutions: for pH< 7 the solution is an acid, for pH > 7 the solution is a base, and for pH = 7 the solution is neutral.

To determine the [H⁺] and the [OH⁻], let's do the dissolution reactions.

a. N₂H₂ → 2N⁻ + 2H⁺

For the stoichiometry (1:2:2), [H⁺] = 2*0.10 = 0.20 M

pH = -log(0.20) = 0.70

b. Ba(OH)₂ → Ba⁺ + 2OH⁻

For the stoichiometry (1:1:2), [OH⁻] = 2*0.10 = 0.20 M

pOH = -log(0.20) = 0.70

pH = 14 - 0.70 = 13.30

c. HOCl → H⁺ + OCl⁻

For the stoichiometry (1:1:1), [H⁺] = 0.10 M

pH = -log(0.10) = 1.00

d. NaOH → Na⁺ + OH⁻

For the stoichiometry (1:1:1), [OH⁻] = 0.10 M

pOH = -log(0.10) = 1

pH = 14 - 1 = 13.00

e. HCl → H⁺ + Cl⁻

For the stoichiometry (1:1:1), [H⁺] = 0.10 M

pH = -log(0.10) = 1.00

Ba(OH)₂ > NaOH > HCl = HOCl >  N₂H₂